Quasi-coherent sheaf which is a vector bundle on curves. This question is inspired by this question. Given a quasi-coherent sheaf on a smooth variety $X$ such that its restrictions to curves are finite dimensional vector bundles. Does it follow that the sheaf is necessarily a finite dimensional vector bundle?
 A: In the case where $\mathscr E$ is coherent, you can re-use the arguments in the answer to the question you cited. First of all, there is an equivalence of categories between the category of vector bundles over $X$ (i.e. algebraic varieties $E$ with a projection $E \to X$ which locally on $X$ looks like the projection $X \times \mathbb A^n \to X$) and the category of locally free sheaves over $X$, the equivalence being given by sending a vector bundle to its sheaf of sections, and in the other direction, using the relative spec of the symmetric algebra of your locally free sheaf to build an (algebraic) vector bundle out of it. This equivalence respects all sorts of things (tensor products, hom-sheaves, pullbacks, etc.) so you can re-formulate the question as:

Given a quasi-coherent sheaf $\mathscr E$ such that its restriction to curves are locally free of finite constant rank, is $\mathscr E$ locally free of finite constant rank?

If you assume $X$ integral and smooth, then all you have to show is that $\mathscr E$ is such that the function $x \mapsto \dim \mathscr E_x/\mathfrak m_x \mathscr E_x$ is constant on the set of closed points of $X$.
If you pass a smooth connected curve between any two closed points $x,y \in X$, as in the argument in the answer of the question you quoted, $\mathscr E|_C$ is locally free of constant rank (because smoothness implies the curve is irreducible), so since any two points can be joined by such a curve, $x \mapsto \dim \mathscr E_x/\mathfrak m_x \mathscr E_x$ is constant.
I am trying to think of a counter-example in the case where $\mathscr E$ is quasi-coherent but not coherent.
Hope that helps,
